Redox II (Edexcel A Level Chemistry): Flashcards

True.

A more positive standard electrode potential indicates a greater tendency to accept electrons and undergo reduction. For example, Br₂ (E^ꝋ = +1.09 V) is more readily reduced than H⁺ (E^ꝋ = 0.00 V).

A high resistance voltmeter ensures that no current flows, so the maximum potential difference is recorded without disturbing the equilibrium positions of the half-cells.

The standard hydrogen electrode is a reference half-cell consisting of hydrogen gas (100 kPa) in equilibrium with H⁺ ions (1.00 mol dm⁻³) at 298 K, using an inert platinum electrode. It is assigned E^ꝋ = 0.00 V.

Platinum is inert — it does not react with the hydrogen gas or H⁺ ions. It provides electrical contact and a surface on which the redox equilibrium is established.

False.

The standard hydrogen electrode is assigned E^ꝋ = 0.00 V. It is the reference point against which all other electrode potentials are compared.

Electrode potential depends on temperature, pressure of any gases, and the concentration of reagents — so standard conditions (298 K, 100 kPa, 1.00 mol dm⁻³) must be used for fair comparison.

A salt bridge is a device containing mobile ions (commonly KNO₃ or KCl solution) that completes the circuit between two half-cells without mixing the solutions.

The standard cell potential is calculated using: E^ꝋ_cell = E^ꝋ_right − E^ꝋ_left.

1. Metal / metal ion half-cell (e.g. Ag⁺ / Ag)

2. Non-metal / non-metal ion half-cell (e.g. Br₂ / Br⁻)

3. Ion / ion half-cell — ions of the same element in different oxidation states (e.g. MnO₄⁻ / Mn²⁺)

False.

Platinum is inert — it does not participate in the reaction. It is used as an electron carrier to make electrical contact with the solution and provide a surface for the redox equilibrium.

Oxidation occurs at the negative pole (more negative E^ꝋ); reduction occurs at the positive pole (more positive E^ꝋ).

Electromotive force (E^ꝋ_cell) is the potential difference (voltage) of an electrochemical cell under standard conditions, calculated as E^ꝋ_right − E^ꝋ_left.

Chlorides and nitrates are generally soluble, so they are unlikely to form precipitates that could disturb the equilibrium positions of the half-cells.

For the cell Cu²⁺/Cu (E^ꝋ = +0.34 V) and Zn²⁺/Zn (E^ꝋ = −0.76 V):

E^ꝋ_cell = (+0.34) − (−0.76) = +1.10 V.

True.

By convention, the more positive half-cell (E^ꝋ_right) is the positive pole and is written on the right; the less positive half-cell (E^ꝋ_left) is the negative pole on the left.

Conventional cell representation is a shorthand notation for describing an electrochemical cell, using | for phase boundaries and ‖ for the salt bridge, without drawing the full cell diagram.

No — conventional cell diagrams are not quantitative. They represent the materials and redox processes involved, not the stoichiometry of the half-equations.

In a cell diagram, a phase boundary is shown by | (single vertical line); the salt bridge is shown by ‖ (double vertical line).

The half-cell with the most negative E^ꝋ is written on the left. Oxidation occurs at the left-hand electrode.

False.

Species in the same phase are separated by a comma. A vertical line (|) represents a phase boundary between species in different phases.

Both ions are aqueous (same phase), so no solid metal electrode is available. An inert platinum electrode acts as the electron carrier. It appears on the outside of its half-cell in the diagram (e.g. Pt | Fe²⁺(aq), Fe³⁺(aq)).

The conventional representation of the Zn/Zn²⁺ and Cu²⁺/Cu cell is:

Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s).

True.

Chlorine gas and chloride ions are in different phases, so they are separated by |. Platinum appears on the outside as the inert electrode making external electrical contact.

The left-hand side represents oxidation, so Fe²⁺ is written first as it is the species being oxidised to Fe³⁺ by loss of an electron.

Thermodynamic feasibility is the prediction that a reaction is likely to occur based on whether the standard cell potential, E^ꝋ_cell, is positive.

E^ꝋ_cell is calculated by subtracting the E^ꝋ of the oxidation half-cell from the E^ꝋ of the reduction half-cell: E^ꝋ_cell = E^ꝋ_reduction − E^ꝋ_oxidation.

True.

A positive E^ꝋ_cell indicates the forward reaction is feasible. The backward reaction, with a negative E^ꝋ_cell, is not feasible under standard conditions.

A larger E^ꝋ_cell corresponds to a bigger change in total entropy; cell potential is also directly proportional to ln *K*.

The two equations are ΔG = −nFE^ꝋ_cell and ΔG = −RT ln K. Since both equal ΔG, it follows that E^ꝋ is directly proportional to ln *K*.

False.

A feasible E^ꝋ_cell gives no information about rate. A reaction may be thermodynamically feasible but kinetically stable if the activation energy is very high or the rate is extremely slow.

1. Concentration changes (non-standard) shift equilibria via Le Chatelier's principle, altering electrode potentials and E^ꝋ_cell.

2. Reaction kinetics — a high activation energy can make the reaction kinetically stable even when E^ꝋ_cell is positive.

If the concentration of the reduced species in a half-cell is greater than 1.0 mol dm^-3, the equilibrium shifts to the left, making the electrode potential more negative.

A kinetically stable reaction is one that is thermodynamically feasible (positive E^ꝋ_cell) but does not proceed at a measurable rate because the activation energy is prohibitively high.

Disproportionation is a redox reaction in which the same element in a single species is simultaneously oxidised and reduced.

The E^ꝋ for Cu²⁺/Cu⁺ (+0.15 V) is less positive than for Cu⁺/Cu (+0.52 V), so Cu⁺ is spontaneously reduced to Cu (s) and oxidised to Cu²⁺ (aq), giving a positive E^ꝋ_cell and a thermodynamically feasible reaction.

In a disproportionation reaction, the half-equation with the more positive E^ꝋ value proceeds as the reduction step, while the half-equation with the more negative E^ꝋ value proceeds as the oxidation step.

True.

E^ꝋ_cell = +0.52 − (+0.15) = +0.37 V, which is positive, so the reaction is thermodynamically feasible.

Oxidation: Cu⁺ (aq) → Cu²⁺ (aq) + e⁻

Reduction: Cu⁺ (aq) + e⁻ → Cu (s)

Overall: 2Cu⁺ (aq) → Cu²⁺ (aq) + Cu (s)

False.

In disproportionation, one species acts as both the oxidising agent and the reducing agent, so only a single reactant is needed.

Assign the half-equation with the more positive E^ꝋ as the reduction step and the other as oxidation. Calculate E^ꝋ_cell = E^ꝋ_reduction − E^ꝋ_oxidation. If the result is positive, the disproportionation is feasible.

For the disproportionation of Cu⁺ (aq): E^ꝋ_cell = +0.52 − (+0.15) = +0.37 V.

A primary cell is a non-rechargeable electrochemical cell in which the redox reactions are irreversible, so it cannot be restored by applying an external voltage.

False.

A battery is correctly used to describe a collection of cells. A car battery, for example, consists of six lead-acid cells joined in series.

A lead-acid battery consists of six cells joined in series, producing a combined voltage of approximately 12 V.

The negative electrode is lead (Pb), the positive electrode is lead(IV) oxide (PbO₂), and the electrolyte is sulfuric acid (H₂SO₄).

A secondary cell is a rechargeable electrochemical cell in which the redox reactions are reversible, so the cell can be restored by applying a voltage greater than the cell voltage.

Lead-acid batteries are very heavy and contain toxic materials (lead and lead(IV) oxide). The sulfuric acid electrolyte is also highly corrosive, creating disposal challenges.

False.

Lithium-ion cells do not suffer from the memory effect, so they can be topped up at any state of charge without loss of capacity.

In a zinc-carbon cell, the zinc casing is the negative electrode, the carbon rod acts as an electron carrier, and the ammonium chloride paste is the electrolyte.

Lithium is used because it has a very low density and a relatively high electrode potential, making the cell lightweight and able to generate a high voltage (3.5–4.0 V).

A fuel cell is an electrochemical cell in which a fuel donates electrons at one electrode and oxygen gains electrons at the other electrode, generating a continuous electrical current.

In an alkaline hydrogen-oxygen fuel cell, the negative electrode half-equation is:

H₂ (g) + 2OH⁻ (aq) → 2H₂O (l) + 2e⁻

The overall equation is the same for both:

2H₂ (g) + O₂ (g) → 2H₂O (l)

The overall cell potential is E^ꝋ = +1.23 V.

False.

A fuel cell operates continuously as long as hydrogen and oxygen are supplied; the energy is not stored in the cell and it does not need recharging.

The hydrogen-oxygen fuel cell is the most common type of fuel cell, in which hydrogen is oxidised at the negative electrode and oxygen is reduced at the positive electrode, producing water as the only product.

1. Water is the only reaction product, so no CO₂ or CO is produced.

2. The reaction occurs at room temperature, so no harmful oxides of nitrogen (NO_x) are formed.

False.

Hydrogen production is currently a by-product of the crude oil industry, meaning it relies on a non-renewable, finite resource, which limits the environmental benefit of fuel cells.

In an acidic hydrogen-oxygen fuel cell, the positive electrode half-equation is:

O₂ (g) + 4H⁺ (aq) + 4e⁻ → 2H₂O (l)

1. Hydrogen is highly flammable, creating safety hazards in production and storage.

2. Hydrogen has a low energy density per unit volume (being a gas), requiring larger and heavier storage containers compared to liquid fuels.

Dilute sulfuric acid is used because it does not react with MnO₄⁻ ions and does not get oxidised under these conditions. Hydrochloric acid would be oxidised by MnO₄⁻, and nitric acid is itself an oxidising agent, both of which would interfere with the titration.

False.

Potassium manganate(VII) titrations are self-indicating. The purple MnO₄⁻ ions are reduced to pale Mn²⁺ ions; a permanent pale pink colour in the flask signals the endpoint.

In a KMnO₄ titration with Fe²⁺ (aq), the molar ratio of MnO₄⁻ : Fe²⁺ is 1 : 5.

Starch is added when the solution is a straw colour. The solution turns blue-black (starch-iodine complex). The endpoint is reached when the blue-black colour disappears, indicating all iodine has reacted.

The overall redox equation for the iodine-thiosulfate titration is:

2S₂O₃²⁻ (aq) + I₂ (aq) → 2I⁻ (aq) + S₄O₆²⁻ (aq)

1. Write the half-equations for oxidant and reductant.

2. Deduce the overall equation and molar ratio.

3. Calculate moles of oxidant used from the titre.

4. Use the ratio to find moles of reductant.

5. Calculate the concentration or percentage purity of the reductant.

True.

The oxidising agent (e.g. ClO⁻) oxidises I⁻ ions to I₂. The iodine produced is then titrated against a known concentration of sodium thiosulfate, allowing the concentration of the oxidising agent to be determined.

MnO₄⁻ (aq) (purple) is reduced to Mn²⁺ (aq) (very pale pink, appearing colourless at low concentration). The half-equation is:

MnO₄⁻ (aq) + 8H⁺ (aq) + 5e⁻ → Mn²⁺ (aq) + 4H₂O (l)